I started by finding the first few powers of each one and attempting to find a pattern. There doesnt seem to be any, and I can't see what else can help me.
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Do you mean $2^{2016}$ etc.? – Angina Seng Jan 01 '18 at 21:37
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1Hint: the number of digits in a number $n$ is $\lfloor\log_{10}(n)\rfloor + 1$. – Mees de Vries Jan 01 '18 at 21:39
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4The product of the two numbers is $10^{2017}$.... – Angina Seng Jan 01 '18 at 21:41
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@MeesdeVries Can you provide a proof as a comment? – Gerard L. Jan 01 '18 at 21:42
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1@GerardL., by induction: all real numbers in $[1, 10)$ have a single decimal digit before the decimal separator, and their 10-logs rounded down are 0; then by induction on $n$ all real numbers in $[10^n, 10^{n+1})$ have $n + 1$ decimal digits before the decimal separator, and their 10-logs rounded down are $n$. – Mees de Vries Jan 01 '18 at 21:46
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It's a direct result of the definition of decimal notation. – fleablood Jan 01 '18 at 21:59
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1Can you write out your "first few powers of each one" computation? Here is what I get: 25 has 2 digits, 425 has 3 digits, 8125 has 4 digits, 16625 has 5 digits, 323125 has 6 digits... looks like a pattern to me! – Qiaochu Yuan Jan 01 '18 at 22:12
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@GerardL. Other proof: any number can be represent using decimal representation(e.g. $126=1|2|6$), say the number $a=a_na_{n-1}a_{n-2}\cdots a_0$, as long as $a_n>0$ we changing $a_i$ won't change the number of elements (it will stay $n+1$), so I can set $a_n=1$ and every other $a_i=0$. I am left with $10\cdots0=10^n$. We can find $n: n=\log_{10}(10^n)$. Now because $\log_{10}(x)$ is continuous and $\log_{10}(c)=\log_{10}(b)+1\implies c=10b$ we can conclude that $\lfloor \log_{10}(a)\rfloor=n$, hence we left with $n+1=\lfloor\log_{10}(a)\rfloor+1$ for any integer $a$ – ℋolo Jan 01 '18 at 22:13
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i think it’s a shame this doesn’t have more upvotes (at the moment, only mine), if only for the joke alone. – Stephen Jan 01 '18 at 23:14
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As indicated in the comments, the number of digits of $x$ is $\lfloor \log_{10}(x) \rfloor+1$. So the number of digits of the number you want is $$\lfloor \log_{10}(2^{2017}) \rfloor + \lfloor \log_{10}(5^{2017})\rfloor+2=\lfloor 2017 \cdot \log_{10}(2) \rfloor + \lfloor 2017\cdot \log_{10}(5)\rfloor+2.$$ Note that the logarithms to base $10$ of both $2$ and $5$ are irrational, and that if $x+y$ is an integer but neither $x$ nor $y$ is then $$\lfloor x \rfloor+ \lfloor y \rfloor=x+y-1.$$ Thus the number of digits you want is $2018$.
Happy New Year! (...and +1).
Stephen
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For the relationship between the number of digits and the logarithm, let $$n=\sum_{r=0}^ma_r10^r$$ with $0\le a_r \le 9$ and $a_m\neq 0$ so that $10^m\le n \lt 10^{m+1}$ and $n$ has $m+1$ digits. Then $$n=10^{m+1} \cdot \frac n{10^{m+1}}$$ and $0.1\le \frac n{10^{m+1}}\lt 1$ so that $\log_{10} n =m+1+e$ where $-1\le e \lt 0$. Taking the floor of this gives $m$ and adding $1$ gives $m+1$ as required.
Mark Bennet
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By definition, any natural number $n$ will have $k+1$ digits if and only if $10^{k}\le n < 10^{k+1}$.[$*$]
Therefore then number $n$ will have $\lfloor \log_{10} n \rfloor +1$ digits.[$**$].
$2^{2017}= 10^{2017 \log_{10} 2}\approx 10^{2017 * 0.30102999566398119521373889472449} \approx 10^{607.1775012542500707461113506593}$ so $10^{607} < 2^{2017} < 10^{608}$.
So $2^{2017}$ has $608$ digits.
Likewise $5^{2017} = 10^{2017\log_{10} 5} \approx 10^{2017*0.69897000433601880478626110527551} \approx 10^{1409.8224987457499292538886493407}$ so $10^{1409} < 5^{2017} < 10^{1410}$
So $2^{2017}$ has $1410$ digits.
So if you wrote them back to back there would be $608+1410 = 2018$ digits.
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[$*$] (That is because all natural numbers can be written uniquely as $n=\sum_{m=0}^k d_m*10^m$ where $d_m \in \{0,1,.....,9\}$ are the digits of $n$ and $d_k \ne 0$. Therefore $n = 10^{k} + (d_m -1)*10^k + \sum_{m=0}^{k-1} d_m*10^m$ and $0\le (d_m-1)*10^k + \sum_{m=0}^{k-1} d_m*10^m \le 899999....... < 9*10^k$)
[$**$] $10^k \le n < 10^{k+1}$ means
$\log_{10} 10^k \le \log_n < \log_{10} 10^{k+1}$
$k \le \log_n < k+1$ so then number of digits is $k+1$ which is one more than the largest number less than or equal to $\log_{10}$.
....
I don't won't to be the crotchety old man[$***$], but this is a basic fact everybody should know and it should be self-evident and obvious.
[$***$] Well,.... maybe I do a little.... It does have its perks....
Perhaps a little more gentle then my crotchety usual self, but... Watch out that you always round up. It's counter intuitive but start using digits before we get to $10^1$ So $10$ itelf has $2$ digits and $100$ has three digits and so on.
fleablood
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I suppose it’s just that you added one for no reason in the middle of your calculation! So probably no rounding error. – Stephen Jan 01 '18 at 22:29
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That is I got that $2^{2017}$ has $608$ digits (true) and $5^{2017}$ has $1410$ digits (true). And then I claimed $609 + 1410 = 2019$ (also true... but as $608 \ne 609$ completely irrelevent!) Somehow in copying $608$ for earlier, I wrote it as $609$.... Argh... – fleablood Jan 01 '18 at 22:32
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1I'll be honest. The joke went over my head.... happy new years everyone.... – fleablood Jan 01 '18 at 22:37
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1"Just wishing it was one year closer to 2021, I am sure." Why? will you reach drinking age in.... Oh, I see. A fellow American (or a global citizen caught in the fall-out) of what I have taken to referring to "the baboon in the siracha sauce"...(which I plagerize from PZ Myers)... Yes, 2021 can't come soon enough for me in that regard. (But ... sheesh.... I will be old in 2021!!!) – fleablood Jan 01 '18 at 22:58
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